EBK GET READY FOR ORGANIC CHEMISTRY
EBK GET READY FOR ORGANIC CHEMISTRY
2nd Edition
ISBN: 9780321830555
Author: KARTY
Publisher: VST
Question
Book Icon
Chapter 16, Problem 16.58P
Interpretation Introduction

(a)

Interpretation:

The splitting pattern is to be determined for the type of H highlighted in the given molecule.

Concept introduction:

Each chemically distinct proton generates a signal in proton nmr spectroscopy. The signal may be modified, split into a number of lines, by nearby protons. Typically only protons three bonds away produce distinct splitting of the signal of interest. The number of lines, called multiplicity, into which the signal is split, equals n+1, where n is the number of protons three bonds away.

Expert Solution
Check Mark

Answer to Problem 16.58P

The signal for the highlighted H will be a singlet.

Explanation of Solution

The structure of the molecule with the specific H highlighted is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  1

The signal for that H can be split by any protons on adjacent carbons.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  2

There is only one such proton. The proton on the nearest OH group is four bonds away, so it will not have any effect. Only the proton on the left is close enough to have an effect, but it is structurally equivalent to the one of interest. Structurally identical protons do not cause splitting of a signal.

Therefore, the signal of the highlighted proton is not split and appears as a singlet.

Conclusion

The multiplicity of the signal is determined on the basis of the number of structurally distinct protons coupled to it.

Interpretation Introduction

(b)

Interpretation:

The splitting pattern is to be determined for the type of H highlighted in the given molecule.

Concept introduction:

Each chemically distinct proton generates a signal in proton nmr spectroscopy. The signal may be modified, split into a number of lines, by nearby protons. Typically only protons three bonds away produce distinct splitting of the signal of interest. The number of lines, called multiplicity, into which the signal is split, equals n+1, where n is the number of protons three bonds away.

The intensities of the individual lines in such signals are in fixed ratios, given by Pascal’s triangle.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  3

Expert Solution
Check Mark

Answer to Problem 16.58P

The signal for the highlighted proton will appear to be a multiplet, with nine peaks.

Explanation of Solution

The structure of the molecule with the highlighted proton is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  4

The highlighted proton has a total of eight coupled protons, six from the two methyl groups and two from the methylene bridge.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  5

Although the methyl protons and the methylene protons are structurally not exactly equivalent, their coupling constants are similar. Therefore, instead of the complex splitting expected of two distinctly differernt proton groups, they will cause a splitting that resembles the splitting by eight equivalent protons. Therefore, the signal will be split into a multiplet with nine peaks.

Conclusion

The multiplicity of the signal is determined on the basis of the number of structurally distinct protons coupled to it.

Interpretation Introduction

(c)

Interpretation:

The splitting pattern is to be determined for the type of H highlighted in the given molecule.

Concept introduction:

Each chemically distinct proton generates a signal in proton nmr spectroscopy. The signal may be modified, split into a number of lines, by nearby protons. Typically only protons three bonds away produce distinct splitting of the signal of interest. The number of lines, called multiplicity, into which the signal is split, equals n+1, where n is the number of protons three bonds away.

Expert Solution
Check Mark

Answer to Problem 16.58P

The signal of the first highlighted protons, from the end methyl group will be split into a triplet. The signal for the second highlighted type will be split into a quartet.

Explanation of Solution

The molecule and the highlighted protons are

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  6

The protons coupled to each of these are shown in the drawing below:

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  7

The one on the left is coupled to two protons from the methylene group. Therefore, the signal for the first type of protons will be split into a triplet.

The second group is coupled to three protons from the methyl group to its right. Therefore, this signal will be split into a quartet.

Conclusion

The multiplicity of the signal is determined on the basis of the number of structurally distinct protons coupled to it.

Interpretation Introduction

(d)

Interpretation:

The splitting pattern is to be determined for the type of H highlighted in the given molecule.

Concept introduction:

Each chemically distinct proton generates a signal in proton nmr spectroscopy. The signal may be modified, split into a number of lines, by nearby protons. Typically only protons three bonds away produce distinct splitting of the signal of interest. The number of lines, called multiplicity, into which the signal is split, equals n+1, where n is the number of protons three bonds away.

Expert Solution
Check Mark

Answer to Problem 16.58P

The signal for the given proton is split into a doublet.

Explanation of Solution

The structure of the molecule with the highlighted proton is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  8

Only one proton is coupled to the highlighted one. Both are aromatic, benzene protons. However, they are not structurally equivalent as the highlighted proton is three bonds away from the electronegative bromine. The other proton is three bonds away from a highly electronegative oxygen. This difference makes them non-equivalent.

Therefore, the signal of the highlighted proton will be split into a doublet.

Conclusion

The multiplicity of the signal is determined on the basis of the number of structurally distinct protons coupled to it.

Interpretation Introduction

(e)

Interpretation:

The splitting pattern is to be determined for the type of H highlighted in the given molecule.

Concept introduction:

Each chemically distinct proton generates a signal in proton nmr spectroscopy. The signal may be modified, split into a number of lines, by nearby protons. Typically only protons three bonds away produce distinct splitting of the signal of interest. The number of lines, called multiplicity, into which the signal is split, equals n+1, where n is the number of protons three bonds away.

Expert Solution
Check Mark

Answer to Problem 16.58P

The signal of the first highlighted H is a triplet and that of the second is a singlet.

Explanation of Solution

The structure of the molecule with the highlighted protons is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  9

The first of these, in the middle of the chain, has two coupled protons while the second, at the end, has no coupled protons.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  10

Therefore, the signal of the first one will be split into a triplet while that of the second will remain a singlet.

Conclusion

The multiplicity of the signal is determined on the basis of the number of structurally distinct protons coupled to it.

Interpretation Introduction

(f)

Interpretation:

The splitting pattern is to be determined for the type of H highlighted in the given molecule.

Concept introduction:

Each chemically distinct proton generates a signal in proton nmr spectroscopy. The signal may be modified, split into a number of lines, by nearby protons. Typically only protons three bonds away produce distinct splitting of the signal of interest. The number of lines, called multiplicity, into which the signal is split, equals n+1, where n is the number of protons three bonds away.

Expert Solution
Check Mark

Answer to Problem 16.58P

The splitting pattern for the given H will be a doublet of triplets (or a triplet of doublets).

Explanation of Solution

The structure of the molecule with the highlighted H is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  11

There are two protons present at the indicated position. They appear to be structurally equivalent, but they are actually diastereotopic. Each one of them is coupled to two identical protons (Hc), as shown below:

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 16, Problem 16.58P , additional homework tip  12

So signal for each proton (Ha and Hb) is split into a triplet by the two coupled protons (Hc). So effectively, the signal will be a doublet of triplets.

Conclusion

The multiplicity of the signal is determined on the basis of the number of structurally distinct protons coupled to it.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Decide whether these proposed Lewis structures are reasonable. proposed Lewis structure Is the proposed Lewis structure reasonable? Yes. :0: Cl C C1: 0=0: : 0 : : 0 : H C N No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* ☐ Yes. No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* Yes. ☐ No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | * If two or more atoms of the same element don't satisfy the octet rule, just enter the chemical symbol as many times as necessary. For example, if two oxygen atoms don't satisfy the octet rule, enter "0,0".
Draw the Lewis structure for the polyatomic trisulfide anion. Be sure to include all resonance structures that satisfy the octet rule. с [ ] - G
1. Calculate the accurate monoisotopic mass (using all 1H, 12C, 14N, 160 and 35CI) for your product using the table in your lab manual. Don't include the Cl, since you should only have [M+H]*. Compare this to the value you see on the LC-MS printout. How much different are they? 2. There are four isotopic peaks for the [M+H]* ion at m/z 240, 241, 242 and 243. For one point of extra credit, explain what each of these is and why they are present. 3. There is a fragment ion at m/z 184. For one point of extra credit, identify this fragment and confirm by calculating the accurate monoisotopic mass. 4. The UV spectrum is also at the bottom of your printout. For one point of extra credit, look up the UV spectrum of bupropion on Google Images and compare to your spectrum. Do they match? Cite your source. 5. For most of you, there will be a second chromatographic peak whose m/z is 74 (to a round number). For one point of extra credit, see if you can identify this molecule as well and confirm by…

Chapter 16 Solutions

EBK GET READY FOR ORGANIC CHEMISTRY

Ch. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - Prob. 16.29PCh. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35PCh. 16 - Prob. 16.36PCh. 16 - Prob. 16.37PCh. 16 - Prob. 16.38PCh. 16 - Prob. 16.39PCh. 16 - Prob. 16.40PCh. 16 - Prob. 16.41PCh. 16 - Prob. 16.42PCh. 16 - Prob. 16.43PCh. 16 - Prob. 16.44PCh. 16 - Prob. 16.45PCh. 16 - Prob. 16.46PCh. 16 - Prob. 16.47PCh. 16 - Prob. 16.48PCh. 16 - Prob. 16.49PCh. 16 - Prob. 16.50PCh. 16 - Prob. 16.51PCh. 16 - Prob. 16.52PCh. 16 - Prob. 16.53PCh. 16 - Prob. 16.54PCh. 16 - Prob. 16.55PCh. 16 - Prob. 16.56PCh. 16 - Prob. 16.57PCh. 16 - Prob. 16.58PCh. 16 - Prob. 16.59PCh. 16 - Prob. 16.60PCh. 16 - Prob. 16.61PCh. 16 - Prob. 16.62PCh. 16 - Prob. 16.63PCh. 16 - Prob. 16.64PCh. 16 - Prob. 16.65PCh. 16 - Prob. 16.66PCh. 16 - Prob. 16.67PCh. 16 - Prob. 16.68PCh. 16 - Prob. 16.69PCh. 16 - Prob. 16.70PCh. 16 - Prob. 16.71PCh. 16 - Prob. 16.72PCh. 16 - Prob. 16.73PCh. 16 - Prob. 16.74PCh. 16 - Prob. 16.75PCh. 16 - Prob. 16.76PCh. 16 - Prob. 16.77PCh. 16 - Prob. 16.78PCh. 16 - Prob. 16.79PCh. 16 - Prob. 16.80PCh. 16 - Prob. 16.81PCh. 16 - Prob. 16.82PCh. 16 - Prob. 16.83PCh. 16 - Prob. 16.84PCh. 16 - Prob. 16.85PCh. 16 - Prob. 16.86PCh. 16 - Prob. 16.87PCh. 16 - Prob. 16.88PCh. 16 - Prob. 16.89PCh. 16 - Prob. 16.1YTCh. 16 - Prob. 16.2YTCh. 16 - Prob. 16.3YTCh. 16 - Prob. 16.4YTCh. 16 - Prob. 16.5YTCh. 16 - Prob. 16.6YTCh. 16 - Prob. 16.7YTCh. 16 - Prob. 16.8YTCh. 16 - Prob. 16.9YTCh. 16 - Prob. 16.10YTCh. 16 - Prob. 16.11YTCh. 16 - Prob. 16.12YTCh. 16 - Prob. 16.13YTCh. 16 - Prob. 16.14YTCh. 16 - Prob. 16.15YTCh. 16 - Prob. 16.16YTCh. 16 - Prob. 16.17YTCh. 16 - Prob. 16.18YTCh. 16 - Prob. 16.19YTCh. 16 - Prob. 16.20YTCh. 16 - Prob. 16.21YTCh. 16 - Prob. 16.22YTCh. 16 - Prob. 16.23YTCh. 16 - Prob. 16.24YTCh. 16 - Prob. 16.25YTCh. 16 - Prob. 16.26YTCh. 16 - Prob. 16.27YTCh. 16 - Prob. 16.28YTCh. 16 - Prob. 16.29YTCh. 16 - Prob. 16.30YTCh. 16 - Prob. 16.31YTCh. 16 - Prob. 16.32YTCh. 16 - Prob. 16.33YTCh. 16 - Prob. 16.34YT
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY